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AP Calculus BC — Unit 9.4

Defining and Differentiating Vector-Valued Functions

Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)

Unit 9.4 Key Concepts Packet
1) What Is a Vector-Valued Function?
Definition
A vector-valued function assigns a vector to each input value:
\( \displaystyle \mathbf{r}(t)=\langle x(t),y(t)\rangle \) (in 2D) or \( \displaystyle \mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle \) (in 3D).
  • The component functions \(x(t),y(t),z(t)\) are ordinary scalar functions.
  • \(t\) is the parameter (often time in motion problems).
  • The set of points traced by \(\mathbf{r}(t)\) is a parametric curve (often a space curve in 3D).
2) Limits, Continuity, and Component Functions
Component-wise idea
\( \displaystyle \lim_{t\to a}\mathbf{r}(t)=\left\langle \lim_{t\to a}x(t),\ \lim_{t\to a}y(t)\right\rangle \) (and similarly for 3D).
  • \(\mathbf{r}(t)\) is continuous at \(t=a\) if each component is continuous at \(a\).
  • This is why AP problems usually reduce vector questions to component function work.
3) Derivative of a Vector-Valued Function
Definition + key result
\( \displaystyle \mathbf{r}'(t)=\lim_{h\to 0}\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} \), and in practice:
\( \displaystyle \mathbf{r}'(t)=\langle x'(t),y'(t)\rangle \) or \( \displaystyle \mathbf{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle \).
  • Differentiation is component-wise.
  • Vector rules behave like scalar rules: product by constant, sum, etc.
4) Velocity, Speed, and Acceleration
Motion interpretation (most common AP context)
If \(\mathbf{r}(t)\) is position, then
\( \mathbf{v}(t)=\mathbf{r}'(t) \) (velocity), \( \mathbf{a}(t)=\mathbf{r}''(t) \) (acceleration), and speed is \( \|\mathbf{v}(t)\|=\|\mathbf{r}'(t)\| \).
  • Speed is a scalar (magnitude), velocity is a vector (direction included).
  • A particle is at rest when \(\mathbf{v}(t)=\mathbf{0}\).
5) Tangent Line / Tangent Vector at a Point
Tangent vector
At \(t=t_0\), the tangent vector is \(\mathbf{r}'(t_0)\). A vector equation for the tangent line is:
\( \displaystyle \mathbf{L}(t)=\mathbf{r}(t_0)+\mathbf{r}'(t_0)(t-t_0) \).
  • In 2D, you can also compute \(dy/dx=\frac{(dy/dt)}{(dx/dt)}\) when \(dx/dt\neq 0\).
  • In 3D, use the vector line equation (AP-friendly and clean).
6) Calculator Expectations + Common AP Traps
  • AP often asks for velocity and acceleration vectors at a given \(t\).
  • Do not confuse speed (\(\|\mathbf{r}'(t)\|\)) with velocity (\(\mathbf{r}'(t)\)).
  • In 2D, \(dy/dx\) is undefined when \(dx/dt=0\) (vertical tangent scenario).
  • Important: If you ever see raw backslashes instead of formatted math, it means MathJax did not typeset that section—this page forces typesetting on load and when solutions open.
Worked Examples + Notes

Example 1 — Differentiate a vector-valued function

Let \( \mathbf{r}(t)=\langle t^2,\ 3t-1\rangle \). Find \(\mathbf{r}'(t)\).

Step 1: Differentiate each component.
Step 2: \( (t^2)'=2t \) and \( (3t-1)'=3 \).
Conclusion: \( \mathbf{r}'(t)=\langle 2t,\ 3\rangle \).

Example 2 — Velocity and speed

A particle has position \( \mathbf{r}(t)=\langle \cos t,\ \sin t\rangle \). Find velocity, speed, and evaluate the speed at \(t=\pi/4\).

Step 1: Velocity \( \mathbf{v}(t)=\mathbf{r}'(t)=\langle -\sin t,\ \cos t\rangle \).
Step 2: Speed \( \|\mathbf{v}(t)\|=\sqrt{\sin^2 t+\cos^2 t}=1 \).
Conclusion: Speed at \(t=\pi/4\) is \(1\).

Example 3 — Acceleration (second derivative)

Let \( \mathbf{r}(t)=\langle t,\ t^2,\ t^3\rangle \). Find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\), then compute \(\mathbf{a}(2)\).

Step 1: \( \mathbf{v}(t)=\mathbf{r}'(t)=\langle 1,\ 2t,\ 3t^2\rangle \).
Step 2: \( \mathbf{a}(t)=\mathbf{v}'(t)=\langle 0,\ 2,\ 6t\rangle \).
Step 3: \( \mathbf{a}(2)=\langle 0,\ 2,\ 12\rangle \).

Example 4 — Tangent line using vectors

Let \( \mathbf{r}(t)=\langle t^2,\ \ln(t)\rangle \) for \(t>0\). Write a vector equation for the tangent line at \(t=1\).

Step 1: Point on the curve: \( \mathbf{r}(1)=\langle 1,\ 0\rangle \).
Step 2: Tangent vector: \( \mathbf{r}'(t)=\langle 2t,\ 1/t\rangle \Rightarrow \mathbf{r}'(1)=\langle 2,\ 1\rangle \).
Step 3: Tangent line: \( \mathbf{L}(t)=\mathbf{r}(1)+\mathbf{r}'(1)(t-1)=\langle 1,\ 0\rangle+\langle 2,\ 1\rangle (t-1) \).
Note: This page forces MathJax typesetting on load and when solution panels open.
Unit 9.4 Multiple-Choice Practice (36 Questions)
Answer Key